Since, $U : H \to H$ satisfies $\lVert U \rVert \le 1$, then $$\left<(I-U)h,h\right> = \lVert h \rVert^2 - \left< Uh,h \right> \ge \lVert h \rVert^2 (1 - \lVert U \rVert) \ge 0 \text{ for all } h \in H$$ We claim that, $N(I-U) = R(I-U)^{\perp}$
If, $h \in N(I-U)$ then, $\left<(I-U)(h - th'), h - th'\right> \ge 0 \implies t^2\left<(I-U)h' , h'\right> - t\left<(I-U)h' , h\right> \ge 0$ for all $h' \in H$ and $t \in \mathbb{R}$.
It follows that $\left<(I-U)h', h\right> = 0$ for all $h' \in H$, i.e., $N(I-U) \subseteq R(I-U)^{\perp}$.
To see the other way inclusion, similarly if $h \in R(I-U)^{\perp}$, then $\left<(I-U)(th' - h),th' - h\right> \ge 0$ for all $h' \in H$ and $t \in \mathbb{R}$,
Implies $t^2\left<(I-U)h' ,h'\right> - t\left<(I-U) h,h'\right> \ge 0$ for all $t \in \mathbb{R}$.
Hence, $\left<(I-U) h,h'\right> = 0$ for all $h' \in H$. That is $h \in N(I-U)$.
Now, since $N((I-U)^{*}) = R(I-U)^{\perp}$ (this is standard result for adjoint of an operator between Banach spaces), it follows $N(I-U) = N(I - U^{*})$, i.e., $U(h) = h \iff U^{*}(h) = h$.
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